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Let $A$ be an $n\times n$ dimensional Covariance matrix. Then $A$ can be any symmstric, positive-definite $n\times n$ matrix with elements between $0$ and $1$.

My understanding: The number $$ k=(1,1,\cdots,1)A^{-1}(1,1,\cdots,1)^{T} $$ represents the number of independent random variables in the system given by the $n$ random variables $x_1,\cdots,x_n$ modulo the covariance relations $\text{Cov}(x_i,x_j)=a_{i,j}$ in the Covariance matrix $A$.

Question: We can find some covariance matrix $A$ such that $k>n$. Why the number of independent random variables in the system given by the $n$ random variables $x_1,\cdots,x_n$ modulo the covariance relations $\text{Cov}(x_i,x_j)=a_{i,j}$ is larger than $n$? This contradicts the probability interpretation.

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  • $\begingroup$ There are a number of errors here. The elements of the covariance matrix have no such bounds. You probably mean the correlation matrix (as defined under the link you provided). Here the elements are in $[-1,1]$, not necessarily in $[0,1]$. In your interpretation of an expression as the number of independent random variables, it's not clear to me how "modulo the covariance relations" is being used. Do you have a link or reference for this interpretation? $\endgroup$ – joriki Sep 14 '15 at 12:12

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